Question

It is desired to prepare 600 mL of 0.100 normal Ca(OH)2 for use in the reaction: 2 HI + Ca(OH)2 CaI2 + 2 H2O How many grams of Ca(OH)2 are needed?

Answer 1

Normality = gram equivalent of solute / volume in L

Normality = Molarity x number of equivalents

2 HI + Ca(OH)2 --------------> CaI2 + 2 H2O

In this reaction 2 moles of H+ participate in the reaction.

So, equivalents are = 2

Given normality = 0.10 normal

So, 0.10 = (No. of moles of Ca(OH)2 /2) / volume in L

0.10 = 2 x (No. of moles of solute / volume in L)

No. of moles of Solute = 0.05 x 0.600 = 0.03 moles

moles = weight of Ca(OH)2 / molar mass of Ca(OH)2

Weight of Ca(OH)2 needed = 0.03 x 74.09 = 2.2 grams

Weight of Ca(OH)2 needed = 2.2 grams